Simplify; express your answer in exponential form. Assume $x\neq 0, k\neq 0$. $\dfrac{{(x^{-3}k^{-4})^{2}}}{{(x^{-1}k^{-2})^{-1}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(x^{-3}k^{-4})^{2} = (x^{-3})^{2}(k^{-4})^{2}}$ On the left, we have ${x^{-3}}$ to the exponent ${2}$ . Now ${-3 \times 2 = -6}$ , so ${(x^{-3})^{2} = x^{-6}}$ Apply the ideas above to simplify the equation. $\dfrac{{(x^{-3}k^{-4})^{2}}}{{(x^{-1}k^{-2})^{-1}}} = \dfrac{{x^{-6}k^{-8}}}{{xk^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{-6}k^{-8}}}{{xk^{2}}} = \dfrac{{x^{-6}}}{{x}} \cdot \dfrac{{k^{-8}}}{{k^{2}}} = x^{{-6} - {1}} \cdot k^{{-8} - {2}} = x^{-7}k^{-10}$